Suppose that for each i=1,...,k,
[tex]u_i = \sum_{j=1}^nc_i^jv_j[/tex]
Then, for an arbitrary linear combination of the u_i,
[tex]\sum_{i=1}^ka^iu_i=\sum_{i=1}^ka^i\left(\sum_{j=1}^nc_i^jv_j\right)=\sum_{j=1}^n\left(\sum_{i=1}^ka^ic_i^j\right)v_j[/tex]
(a linear combination of the v_j !) This shows that [itex]\mathrm{span}(u_1,\ldots,u_k)\subset \mathrm{span}(v_1,\ldots,v_n)[/itex].
And in the same way, if each v_j can be written as a linear combination of the u_i, we obtain [itex]\mathrm{span}(v_1,\ldots,v_n)\subset \mathrm{span}(u_1,\ldots,u_k)[/itex].
And so in that case, [itex]\mathrm{span}(v_1,\ldots,v_n)= \mathrm{span}(u_1,\ldots,u_k)[/itex].
On the other hand, if for instance, u_i cannot be written as a linear combination of the v_j's, then [itex]\mathrm{span}(v_1,\ldots,v_n)\neq \mathrm{span}(u_1,\ldots,u_k)[/itex] since [itex]u_i\in \mathrm{span}(u_1,\ldots,u_k)[/itex] but [itex]u_i \notin\mathrm{span}(v_1,\ldots,v_n)[/itex].